Let $x=y=z=u=2^{n-3}m^{n-1},v=2^{n-2}m^{n}$ $x^n+y^n+z^n+u^n=4x^n=2^{(n-3)n+2}m^{n(n-1)}=2^{(n-1)(n-2)}m^{n(n-1)}= (2^{n-2}m^{n})^{n-1}=v^{n-1}$

RagvaloD wrote: Let $x=y=z=u=2^{n-3}m^{n-1},v=2^{n-2}m^{n}$ $x^n+y^n+z^n+u^n=4x^n=2^{(n-3)n+2}m^{n(n-1)}=2^{(n-1)(n-2)}m^{n(n-1)}= (2^{n-2}m^{n})^{n-1}=v^{n-1}$ is possible more this $x^n+y^n+z^n+u^n=v^{n}$